b) Ta có: \(B=\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)

\(=\dfrac{2}{4+\sqrt{6+2\sqrt{5}}}\)

\(=\dfrac{2}{4+\sqrt{5}+1}\)

\(=\dfrac{2}{5+\sqrt{5}}=\dfrac{5-\sqrt{5}}{10}\)

a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)

\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[.3]{\left(\sqrt{2}-1\right)^3}\)

\(=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)

b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)

\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)

\(\Leftrightarrow B^3+9B-10=0\)

\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)

\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))

c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)

\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)

\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)

\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)

\(\Rightarrow C=1\)

d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)

\(=\sqrt[3]{3}+\sqrt[3]{2}\)

Vậy...

a) Ta có: \(A=\sqrt{\sqrt{3}+\sqrt{2}}\cdot\sqrt{\sqrt{3}-\sqrt{2}}\)

\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)

\(=\sqrt{3-2}=1\)

b) Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}\)

\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}\)

\(=2\sqrt{3}\)

`A=sqrt{sqrt3+sqrt2}.sqrt{sqrt3-sqrt2}`

`=sqrt{(sqrt3+sqrt2)(sqrt3-sqrt2)}`

`=sqrt{3-2}=1`

`b)B=sqrt{5-2sqrt6}+sqrt{5+2sqrt6}`

`=sqrt{3-2sqrt6+2}+sqrt{3+2sqrt6+2}`

`=sqrt{(sqrt3-sqrt2)^2}+sqrt{(sqrt3+sqrt2)^2}`

`=sqrt3-sqrt2+sqrt3+sqrt2=2sqrt3`

`c)C=3-sqrt{3-sqrt5}`

`=3-sqrt{(6-2sqrt5)/2}`

`=3-sqrt{(sqrt5-1)^2/2}`

`=3-(sqrt5-1)/sqrt2`

`=3-(sqrt{10}-sqrt2)/2`

`=(6-sqrt{10}+sqrt2)/2`

\(a,\dfrac{2}{\sqrt{3}-\sqrt{5}}+\dfrac{3-2\sqrt{3}}{\sqrt{3}-2}\)

\(=\dfrac{5-3}{\sqrt{3}-\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}\)

\(=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\sqrt{3}\)

\(=\sqrt{5}+\sqrt{3}+\sqrt{3}\)

\(=\sqrt{5}+2\sqrt{3}\)

\(b,\dfrac{5-\sqrt{5}}{\sqrt{5}-1}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+3}\)

\(=\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)

\(=\sqrt{5}+\dfrac{5-2\sqrt{15}+3}{5-3}\)

\(=\dfrac{2\sqrt{5}+8-2\sqrt{15}}{2}\)

\(=\dfrac{2\cdot\left(\sqrt{5}+4-\sqrt{15}\right)}{2}\)

\(=\sqrt{5}-\sqrt{15}+4\)

#\(Toru\)

\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)

\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)

\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)

\(=\sqrt{29}\)

b: \(B=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)

\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)

c: \(=\sqrt{5}+\dfrac{1}{2}\cdot2\sqrt{5}+\sqrt{5}=3\sqrt{5}\)

a: Ta có: \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)

b: Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)

\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}\)

\(=\sqrt{2}\)

c: \(\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)

a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)

\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)

\(=4-3\cdot A\)

\(\Leftrightarrow A^3+3A-4=0\)

\(\Leftrightarrow A^3-A+4A-4=0\)

\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)

\(\Leftrightarrow A=1\)